What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m
1 answer:
Complete question:
A taut rope has a mass of 0.123 kg and a length of 3.54 m. What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m if the waves are to travel at 28.0 m/s ?
Answer:
The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.
Explanation:
Velocity = Frequency X wavelength
V = Fλ ⇒ F = V/λ
F = 28/0.6 = 46.67 Hz
Angular frequency (ω) = 2πF = 2π (46.67) = 93.34π rad/s
Average power supplied to the rope will be calculated as follows
where;
ω is the angular frequency
A is the amplitude
V is the velocity
μ is mass per unit length = 0.123/3.54 = 0.0348 kg/m
= 1676.159 watts
The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.
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<h2>Answer: 1.011 eV</h2>
Explanation:
The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.
If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>
This is what Einstein proposed:
Light behaves like a stream of particles called photons with an energy :
(1)
So, the energy of the incident photon must be equal to the sum of the Work function
of the metal and the kinetic energy
of the photoelectron:
(2)
Where is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>
In this case and
So, for the first light source of wavelength , and applying equation (2) we have:
(3)
(4)
Now, substituting (1) in (4):
(5)
Where:
is the Planck constant
is the frequency
Now, the <u>frequency has an inverse relation with the wavelength </u>
:
(6)
Where is the speed of light in vacuum
Substituting (6) in (5):
(7)
Then finding :
(8)
We obtain the wavelength of the first light suorce :
(9)
Now, we are told the second light source has the double the wavelength of the first:
(10)
Then: (11)
Knowing this value we can find :
(12)
(12)
(13)
Knowing the value of and
, and knowing we are working with the same work function, we can finally find the maximum kinetic energy
for this wavelength:
(14)
(15)
This is the maximum kinetic energy for the second light source